Problem: $-10pq - 10q - 8r + 3 = 7q + r - 1$ Solve for $p$.
Solution: Combine constant terms on the right. $-10pq - 10q - 8r + {3} = 7q + r - {1}$ $-10pq - 10q - 8r = 7q + r - {4}$ Combine $r$ terms on the right. $-10pq - 10q - {8r} = 7q + {r} - 4$ $-10pq - 10q = 7q + {9r} - 4$ Combine $q$ terms on the right. $-10pq - {10q} = {7q} + 9r - 4$ $-10pq = {17q} + 9r - 4$ Isolate $p$ $-{10}p{q} = 17q + 9r - 4$ $p = \dfrac{ 17q + 9r - 4 }{ -{10q} }$ Swap the signs so the denominator isn't negative. $p = \dfrac{ -{17}q - {9}r + {4} }{ {10q} }$